Math 561 H
Fall 2011
Homework 1 Solutions
Drew Armstrong
1.
Let
G
be a group. Prove that the identity element of
G
is unique (that is, there is only
one element of
G
satisfying the defining property of an identity element). This justifies our
use of the special symbol “
e
”.
Proof.
Suppose that there exist
e, f
∈
G
such that
ae
=
ea
=
a
for all
a
∈
G
and
bf
=
fb
=
b
for all
b
∈
G
. Letting
a
=
f
and
b
=
e
yields
e
=
ef
=
f
.
2.
Let
G
be a finite group.
(a) Show that there are an odd number of
x
∈
G
such that
x
3
=
e
.
(b) Show that there are an even number of
x
∈
G
such that
x
2
6
=
e
.
(Hint: What is the inverse of
x
n
?)
Proof.
Given
x
∈
G
we first note that (
x
n
)

1
= (
x

1
)
n
(make sure you know why this is
true). To prove part (a) we want to show that the size of the set
S
=
{
x
∈
G
:
x
3
=
e
}
is
odd. First note that we have
x
∈
S
⇒
x

1
∈
S
because inverting both sides of
x
3
=
e
gives
(
x

1
)
3
=
e

1
=
e
. Then substituting
x
→
x

1
in the same argument gives
x

1
∈
S
⇒
x
∈
S
.
Thus the elements of
S
come in pairs (
x, x

1
) with
x
6
=
x

1
and possibly some single elements
with
x
=
x

1
. But
x
∈
S
and
x
=
x

1
together imply that
e
=
x
2
⇒
x
=
x
3
⇒
x
=
e
. Hence
e
∈
S
is the only element of
S
that does not come from a pair (
x, x

1
). We conclude that

S

is odd.
To prove part (b) let
T
=
{
x
∈
G
:
x
2
6
=
e
}
. As above we can see that
x
∈
T
⇔
x

1
∈
T
.
Furthermore, by definition no element of
T
satisfies
x
=
x

1
. Hence the elements of
T
come
in pairs (
x, x

1
) and we conclude that

T

is even.
3.
Let
G
be a finite group.
For all
a, b
∈
G
, show that
ab
and
ba
have the same order as
elements of
G
.
Proof.
We wish to show that the sets
h
ab
i
=
{
(
ab
)
k
:
k
∈
Z
}
and
h
ba
i
=
{
(
ba
)
k
:
k
∈
Z
}
have
the same size. This will be done if we can find a bijection from
h
ba
i
to
h
ab
i
. We claim that
φ
a
from Problem 4 is such a bijection.
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 Spring '11
 Armstrong
 Math, Linear Algebra, Algebra, Det, General linear group

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